TITLE: where are my calculations wrong? Expected value
QUESTION [5 upvotes]: I am given a pdf $f(x) = |x-1|$ for $0 \leq x \leq 2,$ 0 otherwise, and asked to find expected value of $X^{2} + X.$ I simply integrated $(x^{2} + x)|x-1|$ from 0 to 2, checked with mathematica, and got $\frac{5}{2}.$ The answer in the back of the book, however, is $\frac{13}6.$ What am I doing wrong? 
Thanks.

REPLY [2 votes]: Resolved in comments. 
The textbook answer is wrong. The correct answer is what the OP got.
$$
\begin{align*}
\mathbf E [X^2 + X] 
&= \int_{-\infty}^{+\infty} f(x) (x^2+x) ~dx
\\ &= \int_{0}^{2} |x-1| (x^2+x) ~dx
\\ &= \int_{0}^{1} (1-x) (x^2+x) ~dx + \int_{1}^{2} (x-1) (x^2+x) ~dx
\\ &= \int_{0}^{1} (x-x^3) ~dx + \int_{1}^{2} (x^3-x) ~dx
\\ &= \left. \Big(\frac{x^2}{2}-\frac{x^4}{4} \Big)\right|_{0}^1 + \left. \Big(\frac{x^4}{4}-\frac{x^2}{2} \Big) \right|_{1}^2 
\\ &= \Big( \frac{1}{2}-\frac{1}{4} \Big) + \Big(\frac{2^4}{4}-\frac{2^2}{2} \Big) - \Big( \frac{1}{4}-\frac{1}{2} \Big) 
\\ &= \frac52 .
\end{align*}
$$